Trilinears           a(b + c)²/(b + c - a) : b(c + a)²/(c + a - b) : c(a + b)²/(a + b - c)
                                    = a²cos²(B/2 - C/2) : b²cos²(C/2 - A/2) : c²cos²(A/2 - B/2)

Trilinears            h(a,b,c) : h(b,c,a) : h(c,a,b), where h(a,b,c) = [r cos(A/2) + s sin(A/2)]², s = semiperimeter, r = inradius

Barycentrics    a³cos²(B/2 - C/2) : b³cos²(C/2 - A/2) : c³cos²(A/2 - B/2)

Let O(A),O(B),O(C) be the excircles. Apollonius's Problem includes the construction of the circle O tangent to the three excircles and encompassing them. (The circle is called the Apollonius circle.) Let A' = O∩O(A), B'=O∩O(B), C'=O∩O(C). The lines AA',BB',CC' concur in X(181). Yff derived trilinears in 1992.

X(181) is the external center of similitude (or exsimilicenter) of the incircle and Apollonius circle. The internal center is X(1682). (Peter J. C. Moses, 8/22/03)

A proof of the the concurrence of lines AA',BB',CC' follows.
            A = exsimilicenter(incircle, A-excircle)
            A' = exsimilicenter(A-excircle, Apollonius circle)
            Let J = exsimilicenter(incircle, Apollonius circle).
By Monge's theorem, the points A, A', J are collinear. In particular, J lies on line AA', and cyclically, J lies on lines BB' and CC'. Therefore, J = X(181). (Darij Grinberg, Hyacinthos, 7461, 8/10/03)

See also

Clark Kimberling, Shiko Iwata, and Hidetosi Fukagawa, Problem 1091 and Solution, Crux Mathematicorum 13 (1987) 128-129; 217-218. [proposed 1985].

X(181) lies on these lines:
1,970    6,197    8,959    10,12    31,51    42,228    43,57    44,375    55,573    56,386    58,1324    171,511    373,748    553,1463    994,1361    1124,1685    1254,1425    1335,1686    1395,1843    1672,1683    1673,1684    1674,1693    1675,1694    1695,1697

X(181) = isogonal conjugate of X(261)
X(181) = X(872)-cross conjugate of X(1500)
X(181) = crosssum of X(I) and X(J) for these (I,J): (21,333), (86,1444)
X(181) = X(I)-beth conjugate of X(J) for these (I,J): (42,181), (660,181), (756,756)